[proofplan] The hypotheses already provide the decisive symbolic model: the coding map $\pi$ is a homeomorphism from the two-sided binary shift onto $\Lambda$. We first identify the two-sided shift with the product of past and future sequence spaces by the concatenation map $\Theta$. Then we prove directly that the relevant binary sequence spaces and their nonempty finite cylinders are compact, perfect, and totally disconnected. Finally, we restrict the homeomorphism $\pi\circ\Theta$ to finite symbolic cylinders and use the assumed fixed-coordinate descriptions of stable and unstable plaques to obtain the stated local product structure. [/proofplan]

[step:Identify the two-sided shift with past and future coordinates] Define \begin{align*} \Phi:=\pi\circ\Theta:\Sigma^-\times\Sigma^+\to\Lambda. \end{align*} The map $\Theta$ is bijective because every bi-infinite sequence $s\in\{0,1\}^{\mathbb Z}$ has a unique negative part $s^-=(s_i)_{i<0}\in\Sigma^-$ and a unique nonnegative part $s^+=(s_i)_{i\geq 0}\in\Sigma^+$. Its inverse is the coordinate projection map \begin{align*} \Theta^{-1}(s)=((s_i)_{i<0},(s_i)_{i\geq 0}). \end{align*} By the definition of the [product topology](/page/Product%20Topology), a coordinate projection is continuous, and a map into a product space is continuous precisely when all its coordinate maps are continuous. Hence $\Theta$ and $\Theta^{-1}$ are continuous. Since $\pi$ is a homeomorphism by hypothesis, $\Phi=\pi\circ\Theta$ is a homeomorphism from $\Sigma^-\times\Sigma^+$ onto $\Lambda$. [/step]

[step:Prove that binary sequence cylinders are Cantor sets]We use the following explicit metric model for binary sequence spaces. If $I$ is one of $\mathbb Z$, $\mathbb Z_{<0}$, or $\mathbb N_0$, choose an enumeration $I=\{i_1,i_2,\dots\}$. For $s,t\in\{0,1\}^I$, define \begin{align*} \rho_I(s,t):=\sum_{j=1}^{\infty}2^{-j}|s_{i_j}-t_{i_j}|. \end{align*} This metric induces the product topology on $\{0,1\}^I$. Compactness follows by the usual diagonal subsequence argument in this metric. Given any sequence $(s_k)_{k\in\mathbb N}$ in $\{0,1\}^I$, pass to a subsequence on which the coordinate $i_1$ is constant, then to a further subsequence on which $i_2$ is constant, and continue. The diagonal subsequence converges coordinatewise to a sequence $s\in\{0,1\}^I$, and coordinatewise convergence is exactly convergence for $\rho_I$. The space has no isolated points whenever $I$ is infinite. Indeed, let $s\in\{0,1\}^I$ and let $N\in\mathbb N$. Choose an index $i_j\in I$ with $j>N$, and define $t\in\{0,1\}^I$ by $t_{i_j}=1-s_{i_j}$ and $t_i=s_i$ for every $i\neq i_j$. Then $t\neq s$ and $\rho_I(s,t)\leq 2^{-j}<2^{-N}$. The space is totally disconnected. If $s,t\in\{0,1\}^I$ are distinct, choose $i\in I$ with $s_i\neq t_i$. The set \begin{align*} A_i:=\{u\in\{0,1\}^I:u_i=s_i\} \end{align*} is both open and closed in the product topology, contains $s$, and does not contain $t$. Thus any connected subset containing both $s$ and $t$ would be separated by the nonempty disjoint relatively clopen sets determined by $A_i$ and its complement. Therefore every connected subset is a singleton. Now let $E\subset\{0,1\}^I$ be a nonempty finite cylinder, meaning that finitely many coordinates are prescribed and all remaining coordinates are free. The same metric argument applies to $E$ after deleting the finitely many fixed coordinates. Since the remaining index set is infinite in each cylinder used in the theorem, $E$ is compact, has no isolated points, and is totally disconnected. Hence every nonempty past cylinder $\Sigma^-[a_{-m},\dots,a_{-1}]$ and every nonempty future cylinder $\Sigma^+[a_0,\dots,a_n]$ is a [Cantor set](/page/Cantor%20Set).[/step]

[guided]The only topological input needed here is the elementary structure of binary sequence spaces. We make it explicit so that no external [compactness theorem](/theorems/2748) is being hidden. Let $I$ be one of the three countable index sets $\mathbb Z$, $\mathbb Z_{<0}$, or $\mathbb N_0$, and fix an enumeration $I=\{i_1,i_2,\dots\}$. We put a metric on $\{0,1\}^I$ by \begin{align*} \rho_I(s,t):=\sum_{j=1}^{\infty}2^{-j}|s_{i_j}-t_{i_j}|. \end{align*} The summand at index $j$ records whether the two sequences differ in the coordinate $i_j$. The coefficient $2^{-j}$ makes later coordinates less expensive, so convergence for $\rho_I$ means exactly this: for every finite list of coordinates, the sequence eventually agrees with the limit on that finite list. That is precisely the product topology on a binary sequence space. We next prove compactness. Take any sequence $(s_k)_{k\in\mathbb N}$ in $\{0,1\}^I$. Since the first coordinate $s_k(i_1)$ takes only the two values $0$ and $1$, some subsequence is constant in the coordinate $i_1$. From that subsequence, choose a further subsequence constant in the coordinate $i_2$. Continuing inductively, for every $r\in\mathbb N$ we obtain a subsequence that is constant in the first $r$ coordinates $i_1,\dots,i_r$. The diagonal subsequence is eventually constant in each fixed coordinate. Define $s\in\{0,1\}^I$ by taking $s_{i_j}$ to be this eventual value in the coordinate $i_j$. For every $N\in\mathbb N$, the first $N$ coordinates of the diagonal subsequence eventually agree with $s$, and the tail of the metric is bounded by $\sum_{j>N}2^{-j}$. Hence the diagonal subsequence converges to $s$ in $\rho_I$. Thus every sequence has a convergent subsequence, so the [metric space](/page/Metric%20Space) is compact. We now prove that there are no isolated points. Fix $s\in\{0,1\}^I$ and a radius scale $2^{-N}$. Because $I$ is infinite, choose an index $i_j$ with $j>N$. Define $t\in\{0,1\}^I$ by flipping only this coordinate: \begin{align*} t_{i_j}:=1-s_{i_j},\qquad t_i:=s_i\text{ for every }i\neq i_j. \end{align*} Then $t\neq s$, while \begin{align*} \rho_I(s,t)=2^{-j}<2^{-N}. \end{align*} Therefore every neighbourhood of $s$ contains a point different from $s$. Finally, we prove total disconnectedness. Let $s,t\in\{0,1\}^I$ be distinct. Choose a coordinate $i\in I$ such that $s_i\neq t_i$. The coordinate cylinder \begin{align*} A_i:=\{u\in\{0,1\}^I:u_i=s_i\} \end{align*} is open because it specifies one coordinate, and it is closed because its complement specifies the opposite value in the same coordinate. It contains $s$ and excludes $t$. If a connected subset contained both $s$ and $t$, then its intersections with $A_i$ and with the complement of $A_i$ would give a separation. Hence no connected subset can contain two distinct points, which is total disconnectedness. The same proof applies to any nonempty finite cylinder. Prescribing finitely many coordinates only removes finitely many degrees of freedom; the unfixed coordinates still form an infinite binary sequence space. Thus each past cylinder and each future cylinder appearing in the theorem is compact, perfect, and totally disconnected, so each is a Cantor set.[/guided]

[step:Transfer compactness, perfectness, and total disconnectedness to $\Lambda$] By the previous step, $\Sigma^-$ and $\Sigma^+$ are compact. Therefore $\Sigma^-\times\Sigma^+$ is compact in the product topology; explicitly, any sequence in the product has a subsequence whose past coordinates converge, and then a further subsequence whose future coordinates converge. Since $\Phi:\Sigma^-\times\Sigma^+\to\Lambda$ is a homeomorphism, $\Lambda$ is compact. The same homeomorphism transfers perfectness and total disconnectedness. If $x\in\Lambda$, write $x=\Phi(s^-,s^+)$ for a unique pair $(s^-,s^+)\in\Sigma^-\times\Sigma^+$. Since $\Sigma^-\times\Sigma^+$ has no isolated points, every neighbourhood of $(s^-,s^+)$ contains a different point; applying $\Phi$ shows that every neighbourhood of $x$ in $\Lambda$ contains a point of $\Lambda$ different from $x$. Hence $\Lambda$ is perfect. If $K\subset\Lambda$ is connected, then $\Phi^{-1}(K)$ is connected in $\Sigma^-\times\Sigma^+$. The product of two totally disconnected binary sequence spaces is totally disconnected by separation in a coordinate, so $\Phi^{-1}(K)$ is a singleton, and therefore $K$ is a singleton. Hence $\Lambda$ is totally disconnected. [/step]

[step:Restrict the symbolic homeomorphism to finite cylinder neighbourhoods] Fix $m,n\geq 0$ and a word $(a_{-m},\dots,a_n)\in\{0,1\}^{m+n+1}$. Define the two-sided symbolic cylinder \begin{align*} Z[a_{-m},\dots,a_n]:=\{s\in\{0,1\}^{\mathbb Z}:s_i=a_i\text{ for every }-m\leq i\leq n\}. \end{align*} By construction of $\Theta$, its preimage is exactly \begin{align*} \Theta^{-1}(Z[a_{-m},\dots,a_n])=\Sigma^-[a_{-m},\dots,a_{-1}]\times\Sigma^+[a_0,\dots,a_n]. \end{align*} The finite cylinder neighbourhood in $\Lambda$ associated to the word is \begin{align*} \Lambda[a_{-m},\dots,a_n]=\pi(Z[a_{-m},\dots,a_n]). \end{align*} Since $\Phi=\pi\circ\Theta$ is a homeomorphism, its restriction \begin{align*} \Phi\big|_{\Sigma^-[a_{-m},\dots,a_{-1}]\times\Sigma^+[a_0,\dots,a_n]}:\Sigma^-[a_{-m},\dots,a_{-1}]\times\Sigma^+[a_0,\dots,a_n]\to\Lambda[a_{-m},\dots,a_n] \end{align*} is a homeomorphism onto its image. This proves the stated product coordinate description of every finite cylinder neighbourhood in $\Lambda$. [/step]

[step:Read the stable and unstable plaques from fixed symbolic coordinates] By hypothesis, for each fixed $s^+\in\Sigma^+$ the set \begin{align*} \Phi(\Sigma^-\times\{s^+\}) \end{align*} is precisely a local unstable plaque in $\Lambda$, and for each fixed $s^-\in\Sigma^-$ the set \begin{align*} \Phi(\{s^-\}\times\Sigma^+) \end{align*} is precisely a local stable plaque in $\Lambda$. Inside a finite cylinder neighbourhood $\Lambda[a_{-m},\dots,a_n]$, the previous step identifies the neighbourhood with a product of the past Cantor set $\Sigma^-[a_{-m},\dots,a_{-1}]$ and the future Cantor set $\Sigma^+[a_0,\dots,a_n]$. Under this identification, fixing the future coordinate gives the assumed unstable plaque direction, and fixing the past coordinate gives the assumed stable plaque direction. Therefore $\Lambda$ has local product structure by stable and unstable Cantor sets in the symbolic coordinates supplied by $\pi$. [/step]